Answer: A
With the same ligand, Δo rises with metal oxidation state (3+ over 2+) and moves down a group (3d < 4d < 5d) because larger, more diffuse d orbitals overlap more with ligands. Ir(III) is a 5d ion of charge 3+, so [Ir(NHX3)X6]X3+ has the largest Δo.